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coldgirl [10]
4 years ago
6

What is the solution to the inequality X+ (7 + 2x) <-3x

Mathematics
1 answer:
dolphi86 [110]4 years ago
3 0

Answer:

x \:  <  \:- 1   \frac{1}{6}

Step-by-step explanation:

The given inequality is

x+ (7 + 2x) <-3x

Expand parenthesis to get

x+ 7 + 2x <-3x

Group similar terms

x+ 2x +3x<-7

Combine similar terms

6x<-7

Divide both sides by 6

x \:  <  \:  -  \frac{7}{6}

Rewrite as improper fraction

x \:  <  \:- 1   \frac{1}{6}

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I am very stuck on this answer simplify 8/12
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3 years ago
A 25 ft. ladder resting against the side of a building forms a right triangle with the ground. The bottom of the ladder is 7 ft.
ruslelena [56]

Answer:

The correct answer is 24

Step-by-step explanation:

to solve this you will need to use the pathagreom theorum

a^{2}+b^{2}=c^{2}

A= one side lenth

B= the secons side lenth

C= hypotnuse

It is helpfull to draw out the situation

you know that the latter is 25 ft, that is your hypotnuse

you also know that the 7 ft away from the base of the building is one of the side lenths, lets call it side a

so plug the numbers into the equation

7^2 + b^2 = 25 ^2

you leave b^2 alone because that is the side you are trying to find

now square 7 and 25 but leave b^2 alone

49 + b^2 = 625

now subtract 49 from both sides

b^2 = 576

now to get rid of the square of b you have to do the opposite and square root both sides removing the square of the B and giving you the answer of..........

B= 24

Hope this helped!! I tryed to explain it as simpil as possiable

6 0
2 years ago
Read 2 more answers
Six friends go out to lunch and decide to split the bill evenly. The bill comes to $61.56. How much does each person owe?
ddd [48]

Answer:

61.56/6

10.26

Step-by-step explanation:


7 0
3 years ago
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CAN SOMEONE PLEASE PLEASE PLEASE HELP ME, YOU’LL GET FREE EASY POINTS IF YOU GIVE ME THE RIGHT ANSWER !!
bekas [8.4K]

Answer:

  1. reflection across BC
  2. the image of a vertex will coincide with its corresponding vertex
  3. SSS: AB≅GB, AC≅GC, BC≅BC.

Step-by-step explanation:

We want to identify a rigid transformation that maps congruent triangles to one-another, to explain the coincidence of corresponding parts, and to identify the theorems that show congruence.

__

<h3>1.</h3>

Triangles GBC and ABC share side BC. Whatever rigid transformation we use will leave segment BC invariant. Translation and rotation do not do that. The only possible transformation that will leave BC invariant is <em>reflection across line BC</em>.

__

<h3>2.</h3>

In part 3, we show ∆GBC ≅ ∆ABC. That means vertices A and G are corresponding vertices. When we map the congruent figures onto each other, <em>corresponding parts are coincident</em>. That is, vertex G' (the image of vertex G) will coincide with vertex A.

__

<h3>3.</h3>

The markings on the figure show the corresponding parts to be ...

  • side AB and side GB
  • side AC and side GC
  • angle ABC and angle GBC
  • angle BAC and angle BGC

And the reflexive property of congruence tells us BC corresponds to itself:

  • side BC and side BC

There are four available congruence theorems applicable to triangles that are not right triangles

  • SSS -- three pairs of corresponding sides
  • SAS -- two corresponding sides and the angle between
  • ASA -- two corresponding angles and the side between
  • AAS -- two corresponding angles and the side not between

We don't know which of these are in your notes, but we do know that all of them can be used. AAS can be used with two different sides. SAS can be used with two different angles.

SSS

  Corresponding sides are listed above. Here, we list them again:

  AB and GB; AC and GC; BC and BC

SAS

  One use is with AB, BC, and angle ABC corresponding to GB, BC, and angle GBC.

  Another use is with BA, AC, and angle BAC corresponding to BG, GC, and angle BGC.

ASA

  Angles CAB and CBA, side AB corresponding to angles CGB and CBG, side GB.

AAS

  One use is with angles CBA and CAB, side CB corresponding to angles CBG and CGB, side CB.

  Another use is with angles CBA and CAB, side CA corresponding to angles CBG and CGB, side CG.

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2 years ago
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How do you calculate the area and the perimeter of a parallelogram and a triangle?
Mnenie [13.5K]

for parallelograms:

area = b(h)

so basically, the length times the height. just like how you would find the area of a square.

perimeter: just add all of the sides together.

for triangles:

area = 1/2(b) x h

so half the length multiplied by the height. for example, if the length was 4, you have to use half of that. so that means 2 times whatever the height is.

perimeter: again, just add all three sides together

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